Could someone please help me to solve this equation using exact values.
9^(2x)+4(9^x)=21
and also check for extraneous roots.
thank you
this is just a quadratic in 9^x. So, it factors out as
9^(2x) + 4(9^x) - 21 = 0
(9^x + 7)(9^x - 3) = 0
9^x cannot be -7, so you are left with
9^x = 3
x = 1/2
thank you for answering the question! but I am still not understanding it properly. If you don't mind solving it step by step so I understand how its done.
Thank you
let u = 9^x
then, 9^(2x) = (9^x)^2 = u^2
Thus all you have is
u^2+4u = 21
u^2+4u-21 = 0
(u+7)(u-3) = 0
so, u = -7 or u=3
But, u = 9^x ...
To solve the equation 9^(2x) + 4(9^x) = 21 using exact values, we can rewrite it in terms of a common base. Since 9 = 3^2, we have:
(3^2)^(2x) + 4(3^2)^x = 21
Simplifying,
3^(4x) + 4(3^(2x)) = 21
Now, let's make a substitution to solve this equation. If we let y = 3^(2x), we can rewrite the equation as:
y^2 + 4y = 21
Rearranging terms,
y^2 + 4y - 21 = 0
Now we have a quadratic equation. We can solve it by factoring or by using the quadratic formula. Let's use factoring:
(y + 7)(y - 3) = 0
Setting each factor equal to zero,
y + 7 = 0 ---> y = -7
or
y - 3 = 0 ---> y = 3
Now, remember that y = 3^(2x). We can substitute these values of y back into the equation to find the values of x:
For y = -7:
3^(2x) = -7
However, exponential functions are always positive, so there are no solutions for this case.
For y = 3:
3^(2x) = 3
Since the bases are the same, we can equate the exponents:
2x = 1
Dividing both sides by 2, we find:
x = 1/2
Therefore, the solution to the equation 9^(2x) + 4(9^x) = 21 using exact values is x = 1/2.